Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 2.5 CONTINUITY ¤ 57 25. By Theorem 7, the exponential function e −5t and the trigonometric function cos 2πt are continuous on (−∞, ∞). By part 4 of Theorem 4, L(t) =e −5t cos 2πt is continuous on (−∞, ∞). 27. By Theorem 5, the polynomial t 4 − 1 is continuous on (−∞, ∞). By Theorem 7, ln x is continuous on its domain, (0, ∞). By Theorem 9, ln t 4 − 1 is continuous on its domain, which is t | t 4 − 1 > 0 = t | t 4 > 1 = {t ||t| > 1} =(−∞, −1) ∪ (1, ∞) 29. The function y = 1 is discontinuous at x =0because the 1+e1/x left- and right-hand limits at x =0are different. 31. Because we are dealing with root functions, 5+ √ x is continuous on [0, ∞), √ x +5is continuous on [−5, ∞), sothe quotient f(x) = 5+√ x √ 5+x is continuous on [0, ∞). Sincef is continuous at x =4, lim x→4 f(x) =f(4) = 7 3 . 33. Because x 2 − x is continuous on R, the composite function f(x) =e x2 −x is continuous on R, so lim x→1 f(x) =f(1) = e1 − 1 = e 0 =1. x 2 if x
F. 58 ¤ CHAPTER 2 LIMITS AND DERIVATIVES ⎧ x +2 ⎪⎨ if x1 TX.10 f is continuous on (−∞, 0) and (1, ∞) sinceoneachoftheseintervals it is a polynomial; it is continuous on (0, 1) since it is an exponential. Now lim f(x) = lim +2)=2and lim f(x) = lim x→0− x→0−(x ex =1,sof is discontinuous at 0. Sincef(0) = 1, f is x→0 + x→0 + continuous from the right at 0. Also lim f(x) = lim ex = e and lim f(x) = lim − x) =1,sof is discontinuous x→1− x→1 − x→1 + x→1 +(2 at 1. Sincef(1) = e, f is continuous from the left at 1. 41. f(x) = cx 2 +2x if x
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Solução_Calculo_Stewart_6e

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